weakly harmonic implies derivative is too

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I have the following problem

Let $R> 0$, $B(0,R) \subset \mathbb R^n$ a ball with radius $R$ and $w \in H^1(B(0,R))$ weakly harmonic, that is $\forall \varphi \in C^\infty_c(B(0,R)) \, :$ $$\int_{B(0,R)} \nabla w \, \nabla \varphi \, \mathrm d x = 0$$

How can I show, that $\partial_i w$ is also weakly harmonic on $B(0,R)$?

I know that for all test functions there holds $\int_{B(0,R)} \nabla w \, \partial_i \nabla \varphi \, \mathrm d x = 0$, but that doesn't imply $\int_{B(0,R)} \nabla \partial_i w \, \nabla \varphi \, \mathrm d x = 0$ for all $\varphi$.

My problem is to show that the derivative exists, although I don't know if $w \in H^2$.